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How to plot a function $f(x)=\frac{3(4+x)}{3(2-x)-16}$ (say $x \in [-15,15]$ ) with the condition that i want to give different color for each of the following cases

(i) when $\frac{x+4}{3x+10}>0$ and $\frac{x^2+8x+12}{3x+10}>0$

(ii) when $\frac{x+4}{3x+10}>0$ and $\frac{x^2+8x+12}{3x+10}<0$

(iii) when $\frac{x+4}{3x+10}<0$ and $\frac{x^2+8x+12}{3x+10}>0$

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4 Answers 4

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Admittedly, not very different from the naïve answer already posted, just avoided repeating the function definitions.

f[x_] := (3 (x + 4))/(3 (2 - x) - 16);

g[x_] := (x + 4)/(3 x + 10);

h[x_] := (x^2 + 8 x + 12)/(3 x + 10);

a[x_] := (g[x] > 0 && h[x] > 0);
b[x_] := (g[x] > 0 && h[x] < 0);
c[x_] := (g[x] < 0 && h[x] > 0);

Plot[{f[x] && a[x], f[x] && b[x], f[x] && c[x]}, {x, -15, 15}, 
  PlotRange -> {-3, 3}, PlotStyle -> Thickness[.01], Frame -> True, 
  Axes -> False]

enter image description here

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You could use the option ColorFunction with ColorFunctionScaling->False. First your conditions:

cond1[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)>0
cond2[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)<0
cond3[x_] := (x+4)/(3x+10)<0 && (x^2+8x+12)/(3x+10)>0

And your function:

f[x_] := (3(4+x))/(3(2-x)-16)

Then:

Plot[f[x], {x, -15, 15},
    PlotRange -> {All, {-3, 3}},
    ColorFunctionScaling -> False,
    ColorFunction -> Function @ Piecewise[
        {
        {ColorData[97][1], cond1[#]},
        {ColorData[97][2], cond2[#]},
        {ColorData[97][3], cond3[#]}
        },
        ColorData[97][4]
    ]
]

enter image description here

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An alternative way to specify a color function: 

cf = ColorData[97] @
     (1 + {1, 2}.UnitStep[{(# + 4)/(3 # + 10), (#^2 + 8 # + 12)/(3 # + 10)}]) &;

f[x_] := (3 (4 + x))/(3 (2 - x) - 16)
Plot[f[x], {x, -15, 15}, 
  PlotRange -> {All, {-3, 3}}, 
  BaseStyle -> AbsoluteThickness[5], 
  ColorFunction -> cf, 
  ColorFunctionScaling -> False]

enter image description here

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  • $\begingroup$ Congratulation in the 200k! Well eine. $\endgroup$
    – user21
    Jun 22, 2019 at 3:47
  • $\begingroup$ Thank you @user21. $\endgroup$
    – kglr
    Jun 22, 2019 at 3:58
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Naïve solution:

f1[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
f2[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) < 0] := (3 (4 + x))/(3 (2 - x) - 16)
f3[x_] /; And[(x + 4)/(3 x + 10) < 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)

Plot[{f1[x], f2[x], f3[x]}, {x, -15, 15}, PlotRange -> {All, {-10, 10}}]

enter image description here

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